Home › MHT-CET › Physics › Electromagnetic Induction › The self-inductance of a coaxial cable with inne…
The self-inductance of a coaxial cable with inner-cylinder radius $a$ and outer-cylinder radius $b$, length $\ell$, carrying current $I$, is:
A$\dfrac{2\pi\mu_0 \ell}{\ln(b/a)}$
B$\dfrac{\mu_0 \ell}{2\pi}\,\ln(b/a)$
C$\dfrac{\mu_0 \ell}{2\pi}(b - a)$
D$\dfrac{\mu_0 \ell}{4\pi}(b^2 - a^2)$
Answer & Solution
Correct answer: B. $\dfrac{\mu_0 \ell}{2\pi}\,\ln(b/a)$
Between the cylinders $B = \mu_0 I/(2\pi r)$. Integrating energy density over the cylindrical shell volume yields $W = (\mu_0 I^2 \ell)/(4\pi) \cdot \ln(b/a)$. Equating to $\tfrac{1}{2}LI^2$ gives $L = \mu_0 \ell \ln(b/a)/(2\pi)$.
Related questions
When a rod moves perpendicular to a magnetic field at constant velocity, the FREE charges An LC circuit oscillates with angular frequency ω equal to:Maxwell's correction to Ampère's law introduced:A bar magnet is dropped through a vertical conducting ring. The magnet:A coil of inductance 0.5 H carries a current changing at 4 A s⁻¹. The induced EMF is:A magnetic flux through a coil changes from 4 Wb to 1 Wb in 0.3 s. The average EMF inducedEddy currents are reduced in transformer cores by:A transformer works on the principle of: